1Some preliminary analysis

III Modular Forms and L-functions

1.3 Mellin transform and Γ-function
We unfortunately have a bit more analysis to do, which we will use a lot later
on. This is the Mellin transform.
Definition (Mellin transform). Let f : R
>0
C be a function. We define
M(f, s) =
Z
0
y
s
f(y)
dy
y
,
whenever this converges.
We want to think of this as an analytic function of
s
. The following lemma
tells us when we can actually do so
Lemma. Suppose f : R
>0
C is such that
y
N
f(y) 0 as y for all N Z
there exists m such that |y
m
y(f)| is bounded as y 0
Then M(f, s) converges and is an analytic function of s for Re(s) > m.
The conditions say
f
is rapidly decreasing at
and has moderate growth at
0.
Proof. We know that for any 0 < r < R < , the integral
Z
R
r
y
s
f(y)
dy
y
is analytic for all s since f is continuous.
By assumption, we know
R
R
0 as
R
uniformly on compact subsets
of C. So we know
Z
r
y
s
f(y)
dy
y
converges uniformly on compact subsets of C.
On the other hand, the integral
R
r
0
as
r
0 converges uniformly on compact
subsets of
{s C
:
Re
(
s
)
> m}
by the other assumption. So the result
follows.
This transform might seem a bit strange, but we can think of this as an
analytic continuation of the Fourier transform.
Example. Suppose we are in the rather good situation that
Z
0
|f|
dy
y
< .
In practice, this will hardly ever be the case, but this is a good place to start
exploring. In this case, the integral actually converges on
iR
, and equals the
Fourier transform of f L
1
(G) = L
1
(R
×
>0
). Indeed, we find
ˆ
G = {y 7→ y
: σ R}
=
R,
and
dy
y
is just the invariant measure on
G
. So the formula for the Mellin
transform is exactly the formula for the Fourier transform, and we can view the
Mellin transform as an analytic continuation of the Fourier transform.
We now move on to explore properties of the Mellin transform. When we
make a change of variables y αy, by inspection of the formula, we find
Proposition.
M(f(αy), s) = α
s
M(f, s)
for α > 0.
The following is a very important example of the Mellin transform:
Definition function). The Γ function is the Mellin transform of
f(y) = e
y
.
Explicitly, we have
Γ(s) =
Z
0
e
y
y
s
dy
y
.
By general theory, we know this is analytic for Re(s) > 0.
If we just integrate by parts, we find
Γ(s) =
Z
0
e
y
y
s1
dy =
e
y
y
s
s
0
+
1
s
Z
0
e
y
y
s
dy =
1
s
Γ(s + 1).
So we find that
Proposition.
sΓ(s) = Γ(s + 1).
Moreover, we can compute
Γ(1) =
Z
0
e
y
dy = 1.
So we get
Proposition. For an integer n 1, we have
Γ(n) = (n 1)!.
In general, iterating the above formula, we find
Γ(s) =
1
s(s + 1) ···(s + N 1)
Γ(s + N).
Note that the right hand side makes sense for
Re
(
s
)
> N
(except at non-
positive integer points). So this allows us to extend Γ(
s
) to a meromorphic
function on {Re(s) > N}, with simple poles at 0, 1, ··· , 1 N of residues
res
s=1N
Γ(s) =
(1)
N1
(N 1)!
.
Of course, since
N
was arbitrary, we know Γ(
s
) extends to a meromorphic
function on C \ Z
0
.
Here are two facts about the Γ function that we are not going to prove,
because, even if the current experience might suggest otherwise, this is not an
analysis course.
Proposition.
(i) The Weierstrass product: We have
Γ(s)
1
= e
γs
s
Y
n1
1 +
s
n
e
s/n
for all
s C
. In particular, Γ(
s
) is never zero. Here
γ
is the Euler-
Mascheroni constant, given by
γ = lim
n→∞
1 +
1
2
+ ··· +
1
n
log n
.
(ii) Duplication and reflection formulae:
π
1
2
Γ(2s) = 2
2s1
Γ(s
s +
1
2
and
Γ(s)Γ(1 s) =
π
sin πz
.
The main reason why we care about the Mellin transform in this course is
that a lot of Dirichlet series are actually Mellin transforms of some functions.
Suppose we have a Dirichlet series
X
n=1
a
n
n
s
,
where the a
n
grow not too quickly. Then we can write
(2π)
s
Γ(s)
X
n=1
a
n
n
s
=
X
n=1
a
n
(2πn)
s
M(e
y
, s)
=
X
n=1
M(e
2πny
, s)
= M(f, s),
where we set
f(y) =
X
n1
a
n
e
2πny
.
Since we know about the analytic properties of the Γ function, by understanding
M(f, s), we can deduce useful properties about the Dirichlet series itself.